Chapter 19 Solutions

19.1: a) 27/56 = 48.2% of the women feel vulnerable while 46/63 = 73% of the men feel vulnerable.; b) A 95% confidence interval for the difference is given by (.73 - .482) ± 1.96 * .0871, or (.0773, .4187).
19.3: a) There were not enough "successes" at Wahtonka High School. (7 < 10), so the large-sample confidence interval should not be used.; b) 8 drug users and 137 athletes at Wahtonka and 28 users out of 143 athletes at Warrenton.; c) A 95% confidence interval for the difference is given by (.0614, .2134).
19.5 H₀: p_rt= p_nt, the proportion of athletes that abuse drugs under a random testing policy is the same as the proportion who abuse drugs under a no testing policy.
H_a: p_rt<p_nt, the proportion of athletes that abuse drugs under a random testing policy is less than the proportion who abuse drugs under a no testing policy.
The test statistic is -3.53 and the p-value is .0002. The proportion of athletes who abuse drugs is significantly lower among those who are randomly tested than those who are not.
19.7 H₀: p_A= p_E, the proportion of African miners who died was equal to the proportion of European miners who died.
H_a: p_A>p_E, the proportion of African miners who died was greater than the proportion of European miners who died.
The test statistic is .98 and the p-value is .16, so there is not a significant difference between the proportion of African and European miners who died.
19.8: a) A 95% confidence interval for the proportion of all adults who use the internet is (.6091, .6505).; b) A 95% confidence interval for the difference in the proportion of non-users and users who expect business to have Web sites that give product information is (.3693, .4506).
19.9 H₀ p₁-p₂= 0: the proportion of rejected papers without statistical help (p₁) is equal to the proportion of rejected papers with statistical help (p₂). H_a p₁-p₂≠ 0: the proportion of rejected papers without statistical help (p₁) is not equal to the proportion of rejected papers with statistical help (p₂). z = 3.39, with p-value < .0005. There is strong evidence for a difference between rejection rates of papers with no statistical help versus those with statistical help.
19.10 A 95% confidence interval for the proportion of papers submitted that include help from a statistician is (0.527, 0.613).
19.11 A 95% confidence interval for the difference between the proportions of papers rejected without review when a statistician is and is not involved in the research is (.063, .218).
19.14: a) This is an observational study. There was no manipulation of the explanatory variable (city) by the researchers.; b) The proportion of New York female Hispanic drivers who wore their seatbelts was 0.829, while the proportion of Boston female Hispanic drivers who wore their seatbelts was 0.570.
H₀: p_B= p_NY, Boston and New York have the same rate of seatbelt use
H _a: p_B<p_NY, smaller proportion of Boston drivers wear seatbelts.
The test statistic is 5.024, the p-value is less than .0001. There is strong evidence that fewer Hispanic female drivers in Boston wear seatbelts.
19.21 H₀: p_m-p_f=0, there is no difference in failure rates between stores headed by men and stores headed by women.
H_a: p_m-p_f=0, there is a difference in failure rates between stores headed by men and stores headed by women. z = -.388, p-value > .5, so there is not significant difference in the failure rates of stores headed by men and stores headed by women.
19.23 see solution in textbook
19.25: a) The proportion of businesses owned by women that failed is .167. The proportion of business owned by men that failed is .151. The p-value for the test is .713.; b) The proportions are the same, but the p-value is now .0336, so it is significant at the α=.05 level.; c) The 95% confidence interval (plus four method) for part (a) is (-.157, .109). The 95% confidence interval for part (b) is (-.049, -.0013). The increased sample size results in a smaller confidence interval.